Question

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Answer 1

сложим почленно. получим

sinx*siny+cosx*cosy=√3/2

cos(x-y)=√3/2

x-y=±π/6+2πn; n∈Z. (1)

Вычтем  первое из второго уравнения. получим -sinx*siny+cosx*cosy=0

cos(x+y)=0;

х+у=π/2+πк, к∈Z. (2)

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а)Сложим  (1) и (2),

2х=π/2+πк+π/6+2πn; 2х=2π/3+πк+2πn;  х=π/3+πк/2+πn;n∈Z; к∈Z.

у=π/2+πк-π/3-πк/2-πn;n∈Z; к∈Z. у=π/6+πк/2-πn;n∈Z; к∈Z.

б) x-y=-π/6+2πn; n∈Z

х+у=π/2+πк,к∈Z.

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2х=π/3+2πn+πк,к∈Z. n∈Z

х=π/6+πn+πк/2,к∈Z. n∈Z; у=π/2+πк-π/6-πn-πк/2,к∈Z. n∈Z;

у=π/3+πк/2-πn,к∈Z. n∈Z;

Answer 2

$\left\{\begin{matrix}sinx\cdot siny=\frac{\sqrt{3}}{4}\\ cosx\cdot cosy=\frac{\sqrt{3}}{4}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}\frac{1}{2}\left ( cos\left ( x-y \right )-cos\left ( x+y \right ) \right )=\frac{\sqrt{3}}{4}\\ \frac{1}{2}\left ( cos\left ( x-y \right )+cos\left ( x+y \right ) \right )=\frac{\sqrt{3}}{4}\end{matrix}\right.$

$\left\{\begin{matrix}\left ( cos\left ( x-y \right )-cos\left ( x+y \right ) \right )=\frac{\sqrt{3}}{2}\\ \left ( cos\left ( x-y \right )+cos\left ( x+y \right ) \right )=\frac{\sqrt{3}}{2}\end{matrix}\right.\Rightarrow \frac{\sqrt{3}}{2}-cos\left ( x+y \right )-cos\left ( x+y \right )=\frac{\sqrt{3}}{2}\\cos\left ( x+y \right )=0\Rightarrow cos\left ( x-y \right )=\frac{\sqrt{3}}{2}$

$\left\{\begin{matrix}cos\left ( x+y \right )=0\\ cos\left ( x-y \right )=\frac{\sqrt{3}}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}x+y=\frac{\pi}{2}+\pi k\\ x-y=\left \{ \frac{\pi}{6}+2\pi k;\frac{5\pi}{6}+2\pi k \right \}\end{matrix}\right.,k\in \mathbb{Z}$

$\left ( x;y \right )=\left \{ \left ( -\frac{\pi}{3}+\frac{\pi}{2}k;-\frac{\pi}{6}+\frac{\pi}{2}k \right );\left ( -\frac{2\pi}{3}+\frac{\pi}{2}k;-\frac{5\pi}{6}+\frac{\pi}{2}k\right ) \right \},k\in \mathbb{Z}$